The locus of the orthocentre of the triangle formed by the lines $(1+p) x-p y+p(1+p)=0$, $(1+q) x-q…
The locus of the orthocentre of the triangle formed by the lines $(1+p) x-p y+p(1+p)=0$, $(1+q) x-q y+q(1+q)=0$ and $y=0$, where $p \neq q$, is
a hyperbola
a parabola
an ellipse
a straight line
Solution
Given, lines are
$
(1+p) x-p y+p(1+p)=0
$
and $(1+q) x-q y+q(1+q)=0$
On solving Eqs. (i) and (ii), we get
$
C\{p q,(1+p)(1+q)\}
$
$\therefore$ Equation of altitude $C M$ passing through $C$ and perpendicular to $A B$ is
$
x=p q
$
$\because$ Slope of line (ii) is $\left(\frac{1+q}{q}\right)$.
$\therefore$ Slope of altitude $B N$ (as shown in figure) is $\frac{-q}{1+q}$.
$\therefore$ Equation of $B N$ is $y-0=\frac{-q}{1+q}(x+p)$
$
\Rightarrow \quad y=\frac{-q}{(1+q)}(x+p)
$
Let orthocentre of triangle be $H(h, k)$, which is the point of intersection of Eqs. (iii) and (iv)
On solving Eqs. (iii) and (iv), we get $x=p q$ and $y=-p q \Rightarrow h=p q$ and $k=-p q \Rightarrow h+k=0$
$\therefore$ Locus of $H(h, k)$ is $x+y=0$
